Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc- Following reaction takes place- Zn - 2HCl - ZnCl2 - H2What would be the volume of hydrogen gas is liberated at STP when 32-65 g of zinc reacts with HCl-

The correct option is C 11.35 L Balanced reaction is: Zn+2HCl → ZnCl_2+H_2 Molar mass of Zn= 65.4 g/molthus 65.4 g of zinc liberates 1 mol i.e. ...

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Standard XII

Chemistry

Preparation of Hydrogen

Hydrogen gas ...

Question

Hydrogen gas is prepared in the laboratory by reacting dilute $H C l$ with granulated zinc. Following reaction takes place:
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with $H C l$ ?

10.03 L

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11.35 L

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11.57 L

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9.53 L

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Solution

The correct option is C $11.35 L$
Balanced reaction is:
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
Molar mass of $Z n =$ 65.4 g/mol
thus 65.4 g of zinc liberates 1 mol i.e. 22.4 L hydrogen gas at STP
32.65 g of zinc will liberate: $\frac{22.4}{65.4} \times 32.65 = 11.18 L$ of hydrogen gas at STP
Hence, option $B$ is correct.

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Similar questions

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place

$Z n + 2 H C l \to Z n C l_{2} + H_{2}$

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3u

When dilute HCl reacts with Zinc metal, the gas liberated is?

Q. What volume of hydrogen at NTP will be liberated when 3.25 g of zinc completely dissolve in dilute HCl?

[At. mass of Zn = 65]

Q. Assertion :Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Reason: Zinc is above hydrogen in reactivity series.

Q. Zinc reacts with both

H C l

and

N a O H

to liberate

H_{2}

gas. This proves that

Z n

is in nature.

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Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place:Zn+2HCl→ZnCl2+H2What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with HCl?

The correct option is C 11.35 LBalanced reaction is:Zn+2HCl→ZnCl2+H2Molar mass of Zn= 65.4 g/molthus 65.4 g of zinc liberates 1 mol i.e. 22.4 L hydrogen gas at STP32.65 g of zinc will liberate: 22.465.4×32.65=11.18L of hydrogen gas at STPHence, option B is correct.

Hydrogen gas is prepared in the laboratory by reacting dilute $H C l$ with granulated zinc. Following reaction takes place:
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with $H C l$ ?